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Find the first partial derivatives of the function. \(g(s, t)=s^{2} t+s t^{-3}\)

Short Answer

Expert verified
The first partial derivatives of the function are: 1. \(\frac{\partial g(s, t)}{\partial s} = 2st + t^{-3}\) 2. \(\frac{\partial g(s, t)}{\partial t} = s^2 - 3st^{-4}\)

Step by step solution

01

Calculate the partial derivative with respect to s

To find the partial derivative of g(s,t) with respect to s, we'll treat t as a constant and use the power rule. The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\). Apply this rule to calculate the derivative of g(s,t) with respect to s: \(\frac{\partial g(s, t)}{\partial s} = \frac{\partial}{\partial s} (s^2 t + s t^{-3})\) \(\frac{\partial g(s, t)}{\partial s} = 2st + t^{-3}\)
02

Calculate the partial derivative with respect to t

To find the partial derivative of g(s,t) with respect to t, we'll treat s as a constant and use the power rule. Apply this rule to calculate the derivative of g(s,t) with respect to t: \(\frac{\partial g(s, t)}{\partial t} = \frac{\partial}{\partial t} (s^2 t + s t^{-3})\) \(\frac{\partial g(s, t)}{\partial t} = s^2 - 3st^{-4}\) The first partial derivatives of the function are: 1. \(\frac{\partial g(s, t)}{\partial s} = 2st + t^{-3}\) 2. \(\frac{\partial g(s, t)}{\partial t} = s^2 - 3st^{-4}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calculus
Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. Fundamental to calculus is the concept of change and motion, characterizing how quantities evolve over time or in relation to each other.

It's split into two major parts: differential calculus, which deals with the rate of change of quantities, and integral calculus, which deals with the accumulation of quantities. When solving problems in calculus, such as finding the rate of change of a function at a point, we use derivatives, a measure that tells us how a function's value will change as its input changes.
Power Rule for Derivatives
The power rule is a basic technique applied in differential calculus for finding derivatives of functions that are powers of variables. It states that if you have a function of the form \(f(x) = x^n\) where \(n\) is any real number, the derivative of \(f\) with respect to \(x\) is \(f'(x) = nx^{n-1}\).

This rule significantly simplifies the process of differentiation by providing a quick and easy way to differentiate functions that would otherwise require a more tedious application of the limit definition of a derivative.
Multivariable Calculus
Multivariable calculus extends concepts from single-variable calculus to functions of multiple variables. This encompasses both partial derivatives and multiple integrals. It allows us to analyze functions that depend on several inputs and is incredibly useful for understanding phenomena across physics, engineering, and economics.

Here, rather than finding the derivative of a function with respect to a single variable, we investigate how the function changes in various directions, defined by different variables. This requires the use of partial derivatives, which measure a function's rate of change with respect to one variable while holding the others constant.
First Partial Derivatives
First partial derivatives are the fundamental tool in multivariable calculus for finding the rate of change of a function with respect to one variable at a time. When a function, like \(g(s, t)\), depends on more than one variable, the first partial derivative with respect to a particular variable treats all other variables as constants.

In the exercise, the first partial derivative of \(g(s, t)\) with respect to \(s\) treats \(t\) as constant and vice versa. Calculating these derivatives provides insight into how the function \(g\) changes in response to changes in each individual variable. This concept is critical in fields such as economics, physics, and engineering, where variables often influence each other.

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Most popular questions from this chapter

Evaluate the first partial derivatives of the function at the given point. \(f(x, y)=\frac{x+y}{x-y} ;(1,-2)\)

Weston Publishing publishes a deluxe edition and a standard edition of its English language dictionary. Weston's management estimates that the number of deluxe editions demanded is \(x\) copies/day and the number of standard editions demanded is \(y\) copies/day when the unit prices are $$\begin{array}{l}p=20-0.005 x-0.001 y \\\q=15-0.001 x-0.003 y\end{array}$$ dollars, respectively. a. Find the daily total revenue function \(R(x, y)\). b. Find the domain of the function \(R\).

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. \(f(x, y)=x y+\ln x+2 y^{2}\)

Evaluate the first partial derivatives of the function at the given point. \(f(x, y, z)=x^{2} y^{2}+z^{2} ;(1,1,2)\)

The total weekly revenue (in dollars) of the Country Workshop realized in manufacturing and selling its rolltop desks is given by $$R(x, y)=-0.2 x^{2}-0.25 y^{2}-0.2 x y+200 x+160 y$$ where \(x\) denotes the number of finished units and \(y\) denotes the number of unfinished units manufactured and sold each week. The total weekly cost attributable to the manufacture of these desks is given by $$C(x, y)=100 x+70 y+4000$$ dollars. Determine how many finished units and how many unfinished units the company should manufacture each week in order to maximize its profit. What is the maximum profit realizable?

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